The image of the point $(4,-13)$ with respect to the line $5 x+y+6=0$ is

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Aicharya · Accepted Answer

Since, $Q$ is the mid point of $P P^{\prime}$, then $Q=\left(\frac{x_1+4}{2}, \frac{y_1-13}{2}\right)$

Also $Q$ lies on the $A B$, then $5\left(\frac{x_1+4}{2}\right)+\left(\frac{y_1-13}{2}\right)+6=0$ $5 x_1+20+y_1-13+12=0$ $5 x_1+y_1+19=0$ $\ldots$ (i) Since, line $P P^{\prime}$ is perpendicular to $A B$, then (Slope of $A B) \times\left(\right.$ Slope of $\left.P P^{\prime}\right)=-1$ $(-5) \times\left(\frac{y_1+13}{x_1-4}\right)=-1$ $5 y_1+65=x_1-4$ $x_1-5 y_1-69=0$ $\ldots$ (ii) Solving Eqs. (i) and (ii), we get $\begin{array}{r}25 x_1+5 y_1+95=0 \\ x_1-5 y_1-69=0 \\ \hline 26 x_1+26=0\end{array}$ $\Rightarrow \quad x_1=-1$ From Eq.(ii) $5 y_1=-1-69$ $5 y_1=-70$ $y_1=-14$ So, image of the point $p$ with respect to $A B$ is $(-1,-14)$

The image of the point $(4,-13)$ with respect to the line $5 x+y+6=0$ is

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